Question

a group of children, adults, and senior citizens attended three different art exhibits that have different ticket prices for each age group. let c represent the number of children, a represent the number of adults, and s represent the number of senior citizens. the system represents the cost of each type of ticket and the total cost of the tickets for all three exhibits. what are the numbers of children, adults, and seniors citizens that attended these three exhibits? \\(\

$$\begin{cases} 6c + 7a + 2s = 990 \\\\ 3c + 3a + 4s = 530 \\\\ 2c + 4a + 4s = 480 \\end{cases}$$

\\) \
120 children, 30 adults, and 30 senior citizens \
20 children, 40 adults, and 70 senior citizens \
100 children, 50 adults, and 5 senior citizens \
100 children, 50 adults, and 20 senior citizens

Explanation:

Step1: Test the first option (120, 30, 30)

Substitute \( c = 120 \), \( a = 30 \), \( s = 30 \) into the first equation: \( 6(120)+7(30)+2(30)=720 + 210+60 = 990 \) (matches). Second equation: \( 3(120)+3(30)+4(30)=360+90 + 120 = 570
eq530 \) (fails).

Step2: Test the second option (20, 40, 70)

Substitute \( c = 20 \), \( a = 40 \), \( s = 70 \) into the first equation: \( 6(20)+7(40)+2(70)=120+280 + 140 = 540
eq990 \) (fails).

Step3: Test the third option (100, 50, 5)

Substitute \( c = 100 \), \( a = 50 \), \( s = 5 \) into the first equation: \( 6(100)+7(50)+2(5)=600+350 + 10 = 960
eq990 \) (fails).

Step4: Test the fourth option (100, 50, 20)

Substitute \( c = 100 \), \( a = 50 \), \( s = 20 \) into the first equation: \( 6(100)+7(50)+2(20)=600+350 + 40 = 990 \) (matches). Second equation: \( 3(100)+3(50)+4(20)=300+150 + 80 = 530 \) (matches). Third equation: \( 2(100)+4(50)+4(20)=200+200 + 80 = 480 \) (matches).

Answer:

100 children, 50 adults, and 20 senior citizens